I've been struggling to understand the explicit details of the completion of an elliptic curve about the origin, and am desperately confused by the explicit details of the resulting group operation. My understanding of the completion of an elliptic curve is as follows.

The goal is to get from algebraic group to complete topological (algebraic?) group.

  1. Begin with an (algebraic connected 1-dimensional( group $G$ over a commutative ring $R$, whose points are defined by, say, $f(x) = x^3 + ax + b$ (whose group operation is the group operation on $Pic(G)$).

(Another perspective: If $A$ is an algebra over a ring $R$, that corresponds to a map $R \to A$, and dually that's a map, $\text{Spec }A \to \text{Spec }R$. In this sense, 'over R' just means 'with a map to Spec R' on the scheme side of things. An elliptic curve has charts $\text{Spec }R[X, Y]/(Y^2-X^3-aX-b)$, there's a natural map to $\text{Spec }R$ on each of these.)

  1. Pick a point $pt=(\alpha, \beta)$ on $G$.

Edit: "Change of coordinates" is NOT a change of point on the formal group (picking a different "pt"), rather it is a different coordinate chart mapped onto the origin of the formal group.

  1. Expand $f(x)$ as a Taylor series around $(x-\alpha)$.

$$f(x) = f(\alpha) + f'(x)(x-\alpha) + \frac{f'(x)}{2!}(x-\alpha)^2 + \frac{f''(x)}{3!}(x-\alpha)^3 + ... $$

  1. Examine the ring of functions whose Taylor series (expanded about $(x-\alpha)$ agrees with the above series up to the nth term. Two functions are declared equal if their truncated Taylor series expanded around $(x-\alpha)$ are equal.

This gives us equivalence classes of functions indexed by $n$, $[f(x)]_n$, aka the jet of $f$ at the point $pt$.

Is this really a group? If $f_1(\alpha) = f_2(\alpha) = \beta^2$ on $(\alpha, \beta)$ then $f_1(\alpha)−f_2(\alpha)$ is zero on $(\alpha, \beta)$, or at least in the neighborhood of $(\alpha, \beta)$.

  1. We endow this collection of rings (of truncated Taylor series) $R[(x-\alpha)]$, $R[(x-\alpha)]/(x-a)^2 = 0$, $R[(x-\alpha)]/(x-a)^3 = 0$, etc. with the ($x-\alpha$)-adic topology:

$$R[(x-\alpha)]/(x-\alpha = 0) \supset R[(x-\alpha)]/((x-\alpha)^2=0) \supset R[(x-\alpha)]/((x-\alpha)^3 = 0) \supset ...$$

A basis of open neighborhoods of this ring $R[(x-\alpha)]$ is given by the sets of the form $r + (x-\alpha)^n$, for $r \in R$.

  1. This gives us a complete topological group (or ring?). The underlying set of points will of course be the ordinary ring of power series around the point, but the group law must be different. The completion doesn't seem to hold any of the structure of the elliptic curve at all!

Here's my question: What is the group operation on the elliptic curve formal group?

How does the additional law on the formal group relate to the addition law of elliptic integrals?

Is it similar to the group law we define on the p-adics?

How do the finite order points of the elliptic curve get involved with restricting our attention to the origin?

I ask about the p-adics because if first few functions in the taylor series of two functions agree, than the functions are close, i.e., if they are divisible by large powers of $(x-p)$, then they are close.

Edit: My general picture of how all of these vocabulary words fit together is as follows. I am not satisfied with this level of detail. I am trying to understand why we want to complete in the first place, and why the procedure of completion works. enter image description here

  • $\begingroup$ A formal group law gives a group law on the maximal ideal of a complete local ring. The point of defining formal group laws is to study the linear part without using the Lie algebra (because it's trivial). For the third question, this is exactly like the p-adic case. By the way, you can find all this on Silverman's book $\endgroup$
    – user40276
    Commented Jul 31, 2015 at 5:57
  • 1
    $\begingroup$ I think you’re making it much too hard, and also hoping for something that (I think) doesn’t exist. More via private communication. $\endgroup$
    – Lubin
    Commented Jul 31, 2015 at 14:25
  • $\begingroup$ The group law, which I do believe you're asking for (as part of your question), is relatively simple. Namely, since $E/k$ is a group scheme, you have a multiplication map $\mu:E\times_k E\to E$. This maps $(e,e)$ to $e$ (the identity element) and so gives you a map $\mathcal{O}_{E,e}\to \mathcal{O}_{E\times E,(e,e)}$. Passing to the completions, and using the fact that both are smooth, we get a map $k[[T]]\to k[[X,Y]]$. The image of $T$ is the formal group law associated to $E$. $\endgroup$ Commented Jul 31, 2015 at 22:11


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